Problem: You are given N items.
The value of the i-th item (1 \leq i \leq N) is v_i.
Your have to select at least A and at most B of these items.
Under this condition, find the maximum possible arithmetic mean of the values of selected items.
Additionally, find the number of ways to select items so that the mean of the values of selected items is maximized.

Constraints

* 1 \leq N \leq 50
* 1 \leq A,B \leq N
* 1 \leq v_i \leq 10^{15}
* Each v_i is an integer.

Input

The input is given from Standard Input in the following format:


N A B
v_1
v_2
...
v_N


Output

Print two lines.
The first line should contain the maximum possible arithmetic mean of the values of selected items. The output should be considered correct if the absolute or relative error is at most 10^{-6}.
The second line should contain the number of ways to select items so that the mean of the values of selected items is maximized.

Examples

Input

5 2 2
1 2 3 4 5


Output

4.500000
1


Input

4 2 3
10 20 10 10


Output

15.000000
3


Input

5 1 5
1000000000000000 999999999999999 999999999999998 999999999999997 999999999999996


Output

1000000000000000.000000
1

Solution (in JAVA):

Solution:
import java.io.IOException;
import java.io.InputStream;
import java.util.NoSuchElementException;
import java.lang.Math;
import java.math.BigDecimal;

public class Main {
	public static void main(String[] args) {
		FS reader = new FS();
		// write reader
		int N = reader.nextInt();
		int A = reader.nextInt();
		int B = reader.nextInt();
		long[] V = new long[N];
		for (int i=0; i<N; i++) {
		    V[i] = reader.nextLong();
		}
		solver(N, A, B, V);
	}
	
	public static void solver(int N, int A, int B, long[] V) {
		// write logic
		// DP:dpv[i][j] max value when choose i items till item j
		//    dpc[i][j] count of max value pattern
		// order:50*50
		long[][] dpv = new long[N+1][N+1];
		long[][] dpc = new long[N+1][N+1];
		for (int i=0; i<N+1; i++) {
		    for (int j=0; j<N+1; j++) {
		        dpv[i][j] = 0;
		        dpc[i][j] = 0;
		    }
    		dpc[i][0] = 1;
		}

		for (int i=1; i<N+1; i++) {
		    for (int j=1; j<i+1; j++) {
		        if (dpv[i-1][j] < dpv[i-1][j-1]+V[i-1]) {
		            dpc[i][j] = dpc[i-1][j-1];
		        } else if (dpv[i-1][j] > dpv[i-1][j-1]+V[i-1]) {
		            dpc[i][j] = dpc[i-1][j];
		        } else { // (dpv[i-1][j] == dpv[i-1][j-1]+V[i-1])
		            dpc[i][j] = dpc[i-1][j] + dpc[i-1][j-1];
		        }
		        dpv[i][j] = Math.max(dpv[i-1][j], dpv[i-1][j-1]+V[i-1]);
		    }
		}

		BigDecimal ans = new BigDecimal(0);
		long anscnt = 0;
		for (int i=A; i<B+1; i++) {
		    BigDecimal check = new BigDecimal(dpv[N][i]);
		    check = check.divide(new BigDecimal(i), 10, BigDecimal.ROUND_HALF_UP);
		    if (ans.compareTo(check) < 0) {
		        anscnt = dpc[N][i];
		    } else if (ans.compareTo(check) == 0) {
		        anscnt += dpc[N][i];
		    }
		    ans = ans.max(check);
		}

		System.out.println(ans.toPlainString());
		System.out.println(anscnt);
	}
	
	// Read Class
	static class FS {
		private final InputStream in = System.in;
		private final byte[] buffer = new byte[1024];
		private int ptr = 0;
		private int buflen = 0;
		private boolean hasNextByte() {
			if (ptr < buflen) {
				return true;
			} else {
				ptr = 0;
				try {
					buflen = in.read(buffer);
				} catch (IOException e) {
					e.printStackTrace();
				}
				if (buflen <= 0) {
					return false;
				}
			}
			return true;
		}
	
		private int readByte() { return hasNextByte() ? buffer[ptr++] : -1;}
		private boolean isPrintableChar(int c) {return 33 <= c && c <= 126;}
		private void skipUnprintable() {while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;}
	
		public boolean hasNext() { skipUnprintable(); return hasNextByte();}
		public String next() {
			if (!hasNext()) throw new NoSuchElementException();
			StringBuilder sb = new StringBuilder();
			int b = readByte();
			while(isPrintableChar(b)) {
				sb.appendCodePoint(b);
				b = readByte();
			}
			return sb.toString();
		}
		public int nextInt() {
			return (int)nextLong();
		}
		public long nextLong() {
			if (!hasNext()) throw new NoSuchElementException();
			long n = 0;
			boolean minus = false;
			int b = readByte();
			if (b == '-') {
				minus = true;
				b = readByte();
			}
			if (b < '0' || '9' < b) {
				throw new NumberFormatException();
			}
			while(true) {
				if ('0' <= b && b <= '9') {
					n *= 10;
					n += b - '0';
				} else if(b == -1 || !isPrintableChar(b)) {
					return minus ? -n : n;
				} else {
					throw new NumberFormatException();
				}
				b = readByte();
			}
		}
	}
}